University Quantum Mechanics II · unit UQMII-02
The Hydrogen Radial Equation Solved
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Centre of mass, relative coordinates, and the reduced mass: infer an input with uncertainty
The reduced mass μ = m₁m₂/(m₁ + m₂) is the inertia of the separation r₁ − r₂ of two interacting particles: once the free centre-of-mass motion is split off, the pair's internal problem is exactly a one-body problem of mass μ in the potential V(r). Use the supplied topic model Y_out = 1/(1+10^(-q)), where Y_out is REDUCED MASS μ/mₑ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Partner mass, log₁₀(m₂/mₑ), dimensionless. Keep no other adjustable input fixed. For this exercise use only 0.40375 ≤ q ≤ 3.39625 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.99413497, with a bounded reading uncertainty ±0.00081954. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00029925, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
REDUCED MASS μ/mₑ against q
Predictions from the supplied Centre of mass, relative coordinates, and the reduced mass model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): REDUCED MASS μ/mₑ
Horizontal axis (x): q: Partner mass, log₁₀(m₂/mₑ)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Partner mass, log₁₀(m₂/mₑ) | REDUCED MASS μ/mₑ |
|---|---|
| 0.40375 | 0.7170080734221427 |
| 1.151875 | 0.9341520057683236 |
| 1.9 | 0.9875672647455576 |
| 2.648125 | 0.9977566366257081 |
| 3.3962499999999998 | 0.9995986015973534 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(q) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.40375) = 0.717008 and Y_out(3.39625) = 0.999599. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 0.40375 + (0.994135 − 0.717008)(3.39625 − 0.40375)/(0.999599 − 0.717008) = 3.33839. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives q = 2.22918; substituting back gives Y_out = 0.99413497.
- The interpolation error in the input is |q_line − q| = 1.10922. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.99331543 and 0.99495451. Sort the roots even if the response decreases: 2.17201 ≤ q ≤ 2.2949.
- The lower and upper input deviations are 0.0571614 and 0.0657248. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer q ≈ 2.22918; compatible interval [2.17201, 2.2949]. Endpoint interpolation gives 3.33839.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The dimensionless Coulomb radial equation: infer an input with uncertainty
The dimensionless Coulomb radial equation is the radial equation for u = rR with r measured in units of 1/κ = ħ/√(−2μE), which leaves l and one pure number, ρ₀ = 2Z/(κa), as its only parameters, the unknown energy being carried inside ρ₀. Use the supplied topic model Y_out = -54.393/rho0^2, where Y_out is HYDROGEN ENERGY −4ℛ/ρ₀² (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rho0 is Coulomb parameter ρ₀, dimensionless. Keep l (Orbital quantum number l) = 1 fixed. For this exercise use only 1.95625 ≤ rho0 ≤ 9.04375 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -1.3793535 eV, with a bounded reading uncertainty ±0.10715 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rho0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00070875, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
HYDROGEN ENERGY −4ℛ/ρ₀² against rho0
Predictions from the supplied The dimensionless Coulomb radial equation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): HYDROGEN ENERGY −4ℛ/ρ₀² / eV
Horizontal axis (x): rho0: Coulomb parameter ρ₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rho0: Coulomb parameter ρ₀ | HYDROGEN ENERGY −4ℛ/ρ₀² / eV |
|---|---|
| 1.9562499999999998 | -14.21327971092897 |
| 3.7281249999999995 | -3.91347065762913 |
| 5.5 | -1.7981157024793388 |
| 7.271875 | -1.0286087222934965 |
| 9.04375 | -0.665037164325877 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(rho0) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.95625) = -14.2133 eV and Y_out(9.04375) = -0.665037 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives rho0_line = 1.95625 + (-1.37935 − -14.2133)(9.04375 − 1.95625)/(-0.665037 − -14.2133) = 8.67007. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives rho0 = 6.27963; substituting back gives Y_out = -1.3793535 eV.
- The interpolation error in the input is |rho0_line − rho0| = 2.39044. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.4865035 eV and -1.2722035 eV. Sort the roots even if the response decreases: 6.04907 ≤ rho0 ≤ 6.53873.
- The lower and upper input deviations are 0.230556 and 0.259102. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer rho0 ≈ 6.27963; compatible interval [6.04907, 6.53873]. Endpoint interpolation gives 8.67007.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Behaviour at infinity and at the origin, and v(ρ): infer an input with uncertainty
The asymptotic factors of the Coulomb problem are ρ^(l+1), which u follows near the nucleus, and e^(−ρ), which sets its decay far from it up to a power of ρ; every bound state shares them, and dividing both out of u defines v(ρ), which then obeys Kummer's equation in 2ρ exactly. Use the supplied topic model Y_out = -54.39/p0^2, where Y_out is HYDROGEN E = −54.39 eV/ρ₀² (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p0 is Energy parameter ρ₀ = 2/(κa), dimensionless. Keep l (Orbital quantum number l) = 1 fixed. For this exercise use only 1.85 ≤ p0 ≤ 8.15 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -1.6781729 eV, with a bounded reading uncertainty ±0.1289 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
HYDROGEN E = −54.39 eV/ρ₀² against p0
Predictions from the supplied Behaviour at infinity and at the origin, and v(ρ) model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): HYDROGEN E = −54.39 eV/ρ₀² / eV
Horizontal axis (x): p0: Energy parameter ρ₀ = 2/(κa)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p0: Energy parameter ρ₀ = 2/(κa) | HYDROGEN E = −54.39 eV/ρ₀² / eV |
|---|---|
| 1.85 | -15.891891891891891 |
| 3.425 | -4.636581597314722 |
| 5 | -2.1756 |
| 6.574999999999999 | -1.2581358701152254 |
| 8.149999999999999 | -0.8188490345891832 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(p0) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.85) = -15.8919 eV and Y_out(8.15) = -0.818849 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p0_line = 1.85 + (-1.67817 − -15.8919)(8.15 − 1.85)/(-0.818849 − -15.8919) = 7.79083. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives p0 = 5.693; substituting back gives Y_out = -1.6781729 eV.
- The interpolation error in the input is |p0_line − p0| = 2.09783. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.8070729 eV and -1.5492729 eV. Sort the roots even if the response decreases: 5.4862 ≤ p0 ≤ 5.9251.
- The lower and upper input deviations are 0.206799 and 0.232098. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer p0 ≈ 5.693; compatible interval [5.4862, 5.9251]. Endpoint interpolation gives 7.79083.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The power series for v(ρ) and its recursion: infer an input with uncertainty
The Coulomb recursion is the rule cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)] that fixes every coefficient of the power series v(ρ) = Σ cⱼρʲ from the first one, c₀, at any value of the energy parameter ρ₀. Use the supplied topic model Y_out = (2*(j+l+1)-rho0)/((j+1)*(j+2*l+2)), where Y_out is RATIO cⱼ₊₁/cⱼ AT j (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rho0 is Energy parameter ρ₀, dimensionless. Keep l (Orbital quantum number l) = 0; j (Coefficient index j) = 20 fixed. For this exercise use only 1.275 ≤ rho0 ≤ 10.725 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.075672078, with a bounded reading uncertainty ±0.0011966. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rho0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000945, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RATIO cⱼ₊₁/cⱼ AT j against rho0
Predictions from the supplied The power series for v(ρ) and its recursion model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RATIO cⱼ₊₁/cⱼ AT j
Horizontal axis (x): rho0: Energy parameter ρ₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rho0: Energy parameter ρ₀ | RATIO cⱼ₊₁/cⱼ AT j |
|---|---|
| 1.275 | 0.08814935064935066 |
| 3.6374999999999997 | 0.08303571428571428 |
| 6 | 0.07792207792207792 |
| 8.362499999999999 | 0.07280844155844156 |
| 10.725 | 0.0676948051948052 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(rho0) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.275) = 0.0881494 and Y_out(10.725) = 0.0676948. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rho0_line = 1.275 + (0.0756721 − 0.0881494)(10.725 − 1.275)/(0.0676948 − 0.0881494) = 7.0395. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives rho0 = 7.0395; substituting back gives Y_out = 0.075672078.
- The interpolation error in the input is |rho0_line − rho0| = 5.32907e-15. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.074475478 and 0.076868678. Sort the roots even if the response decreases: 6.48667 ≤ rho0 ≤ 7.59233.
- The lower and upper input deviations are 0.552829 and 0.552829. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer rho0 ≈ 7.0395; compatible interval [6.48667, 7.59233]. Endpoint interpolation gives 7.0395.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.